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Theorem

Let $ A $ be a set in S. Then
A ∩ Ø = Ø



Proof

Let x ∈ S, where S is the universal set.

First we show that if A ∩ Ø ⊂ Ø.
We know this is true because the set resulting from the union of two sets is a subset of both of the sets (proof).

Next, we want to show that A ∩ Ø ⊂ Ø.
Let x ∈ Ø. The antecedent (i.e. the "if") part is false by definition of the empty set. Then x ∈ Ø ⇒ x ∈ (A ∩ Ø) is true and we have that Ø ⊂ A ∩ Ø.

Since A ∩ Ø ⊂ Ø and Ø ⊂ A ∩ Ø, we have that A ∩ Ø = Ø.
$ \blacksquare $



References



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